Question: Find $\lim_{h\to 0}\dfrac{6\sqrt{9+h}-6\sqrt{9}}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $6$ (Choice C) C $18$ (Choice D) D The limit doesn't exist
Solution: The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $6\sqrt{9+h}-6\sqrt{9}$, we can tell that the function is $f(x)=6\sqrt{x}$ and the $x$ -value is $9$. In other words, the limit expression is equal to $f'(9)$ for $f(x)=6\sqrt{x}$. Let's find $f'(x)$ : $f'(x)=6\cdot \dfrac12{x^{^{{-\scriptsize\dfrac12}}}}=\dfrac{3}{\sqrt{x}}$ Now let's evaluate $f'(9)$ : $f'(9)=\dfrac{3}{\sqrt{9}}=1$ In conclusion, $\lim_{h\to 0}\dfrac{6\sqrt{9+h}-6\sqrt{9}}{h}=1$.